坐标系的平移不改变矢量的分量？曲线坐标系中
”It is worthwhile to mention that vectors and tensors themselves remain invariantupon a change of basis - they are said to be independent of any coordinate system.
However, their respective con1ponents do depend upon the coordinate system introduced,
which is arbitrary. The components change their magnitudes by a rotation
(and/or reflection) of the basis vectors, but are independent of any translation“ 引自 Holzapfel 《Nonlinear Solid l\fechanics ： A Continuum Approach for Engineering》
这最后一句话怎么理解？
直角坐标系是显然的，曲线坐标系中也成立？
这里提到了任意的坐标系，所以无所谓直角坐标还是曲线坐标。如果列出基向量的平动变换矩阵的话，应该能够看出来基向量的平动不影响张量的分量。 这个课件中也有类似的描述：
http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/Chapter_1_Vectors_Tensors/Vectors_Tensors_05_Coordinate_Transformation_Vectors.pdf
tonnyw 发表于 2014-11-2 11:14
这里提到了任意的坐标系，所以无所谓直角坐标还是曲线坐标。如果列出基向量的平动变换矩阵的话，应该能够看 ...
多谢，我想我弄清楚了。我把直线坐标系的平移 和 曲线坐标系中的 “平移”混淆了。
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